This is a community wiki answer based on the comments, plus some remarks of my own.
In principle, yes, you could proceed by extending the $p$-adic valuation from $\mathbb Q$ to $\overline{\mathbb Q}$, and the extending from $\overline{\mathbb Q}$ to $\mathbb C$. But you have to be careful if you want to go this route. In particular, there is not a unique way to extend the $p$-adic valuation from $\mathbb Q$ to $\overline{\mathbb Q}$.
In the comments, it was suggested that perhaps one could define
$$|\alpha|_p := |N_{{\mathbb Q}(\alpha)/{\mathbb Q}}(\alpha)|_p^{1/[{\mathbb Q}(\alpha):{\mathbb Q}]}.$$
So for example, this would give $|1+2i|_5=|1−2i|_5=|5|^{1/2}_5=5^{−1/2}$. However,
then we deduce that
$$|(1+2i)+(1−2i)|_5 = |2|_5 = 1 > 5^{-1/2} = \max(|1+2i|_5,|1−2i|_5),$$
contradicting the ultrametric inequality.
In general, the number of extensions of $|\cdot|_p$ from $\mathbb Q$ to a finite extension $K$ is equal to the number of prime ideals in ${\cal O}_K$ that lie over $p$. The two extensions of $|\cdot|_5$ to ${\mathbb Q}(i)$ are related to the two primes $1+2i$ and $1−2i$ that divide $5$. For one of the extensions, $|1+2i|=1/5$ (not $1/\sqrt{5}$) and $|1−2i|=1$, while for the other extension $|1−2i|=1/5$ and $|1+2i|=1$.
However, the lack of uniqueness is not a dealbreaker; for every finite extension of $\mathbb Q$ you can pick an extension of $|\cdot|_p$, and use Zorn's lemma to ensure that you're making choices that are consistent with each other.
Once you have an extension to all algebraic numbers, you can extend to transcendental numbers $t$ by the so-called Gauss norm construction:
$$ |a_n t^n+\cdots +a_1 t + a_0| := \max_i |a_i|$$
and extending to rational functions of $t$ by multiplicativity.
Although this works, it is generally considered less natural than the standard approach because the extension of the $p$-adic norm from $\mathbb{Q}_p$ to $\overline{\mathbb{Q}}_p$ is unique. This uniqueness is one way in which local fields are simpler than global fields.
Best Answer
You can start with $\mathbb{Z}$ equipped with the $p$-adic norm. Then consider its completion as a metric space, which is $\mathbb{Z}_p$. Then you can consider $\mathbb{Z}_p$ as a ring extending $\mathbb{Z}$. Finally, the field of fractions of $\mathbb{Z}_p$ will be $\mathbb{Q}_p$ and the topologies in these algebraic structures are always induced by an extension of the $p$-adic norm on $\mathbb{Z}$. Since $\mathbb{Z}_p$ with the $p$-adic norm is already complete (indeed, compact), you cannot build $\mathbb{Q}_p$ from it by using only topology.